Question

prove that √2,√5,√3,√7 are irrational numbers​

Answer 1

Step-by-step explanation:

root 2

let us assume that root of 2 is an rational no and root 2=a/b were a and b are co primes

b root 2 = a

sq. on both sides

2b²=a²- equation 1

now 2 divided a²

and 2 divides a

now let a=2c were c is integer

substitute a = 2c in equation 1

2b²=(2c)²

2b²= 4c²

b²=2c² -eq. 2

now 2 divides b² as well as b

from 1 and 2 we observe that root2 is an irr. no.

this contradics our assumption.

our assumption is wrong.

same way do for every question substitute

3 or or5 or 7 instead of 2. thank you

Answer 2

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let's suppose √2 is a rational number.

NOW, let √2 = P/Q , where p and q are coprime number and Q ≠ 0 . [BECAUSE RATIONAL NO. IS IN P/Q FORM.]

SQUARING ON BOTH SIDE.

(√2)² = (P/Q)²

2 = P²/Q²

2Q² = P²

Q² = P²/2

IT MEANS THAT P² IS DIVISIBLE BY 2.

ALSO P IS DIVISIBLE BY 2.

let P = 2m FOR SOME INTEGERS.

(√2)² = (2M)²/Q²

2 = 4M²/Q²

Q² =4M²/2

Q² = 2M²

Q²/2 = M²

IT MEANS THAT Q² IS ALSO DIVISIBLE BY 2.

ALSO Q IS DIVISIBLE BY 2.

BUT, IN A CO-PRIME NUMBER THE VARIABLES ARE ONLY DIVISIBLE BY 1.

SO, IT MEANS THAT OUR ASSUMPTION √2 IS A RATIONAL NO. IS WRONG.

IT MEANS THAT IF √2 IS NOT A RATIONAL NO. THEN IT'S AN IRRATIONAL NO.

HENCE, PROVED.

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